Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.45. Calculate the Ka for the acid.

Answer 1

Using pH to determine [H+]
10^-pH= 10^-2.45 = 2.1 x 10^-3 

Now,
Ka=[H=][A-]/[HA] 

So,
(2.1 x 10^-3)^2/0.0136 - (2.1 x 10^-3) 
Ka=2.5 x 10^4

Answer 2

Answer: $K_a=1.2\times 10^{-6}$

Explanation: $HA\rightarrow H^++A^-$

initially conc.         c                              0         0

At eqm.    $c(1-\alpha)$      $c\alpha$        $c\alpha$

The expression for dissociation constant is,

$K_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}$

And,

$[H^+]=c\alpha$

$pH=-log[H^+]=2.45$

$H^+=3.5\times 10^{-3}$

$3.5\times 10^{-3}=0.0136\alpha$

$\alpha=0.26$

Now put all the given values in this expression, we get

$K_a=\frac{(0.0136\times 0.26)^2}{0.0136(1-0.26)}$

$K_a=1.2\times 10^{-6}$